Projectile Motion at an Angle: Solving for Velocity, Time, and Height

In summary: The velocity at max height is just the magnitude of the horizontal component, 7.5 m/s horizontal direction.
  • #1
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Homework Statement


You throw a ball. The ball leaves your hand with a velocity of 15 m/s at an angle 60 degrees above the horizontal. Acceleration due to gravity is 9.8 m/s/s.
a) Determine the velocity (magnitude and direction) of the ball at the maximum height.
b) How much time does it take for the ball to reach maximum height?
c) Determine the maximum height.

Homework Equations


(I know how to draw the diagram, so that part is good)
I think the components of the velocity at maximum height can be found using these equations:
a) VoY= Vosin60 = 15m/s(sin60) = 12.99 m/s velocity of ball in vertical direction (initial)
VoX= Vocos60 = 15 m/s(sin60) = 7.5 m/s velocity of ball in horizontal direction (initial)
Since there is no vertical velocity at max height, then the velocity at max height is just the magnitude of the horizontal component: 7.5 m/s horizontal direction?

b) Time = Vy= VoY + ayt , set Vy = 0 at Y max.
0 = 12 m/s + (-9.8 m/s/s)t
t = (Vfy - Voy)/ -9.8m/s/s = (0 - 12.99)/-9.8 m/s/s = 1.32 seconds

c) Ymax. = Voyt + 1/2at^2 = (12.99 m/s)(1.32 sec.) + 1/2(-9.8m/s/s)(1.32 sec.)^2

= 8.60 meters max height


The Attempt at a Solution


see above. I am not sure if i did this problem correctly, or if i understand it quite. there are some things about it that i understand, others not so much.
I think the velocity at max height would just be the component 7.5 m/s in the horizontal direction, since suposedly, there is no vertical velocity at max height. Is this assumption correct? Have i used the correct equations and in the right context?? If not, please help me to understand this type of problem better.
I understand that at max height, there is only the force of gravity 9.8 m/s/s downward, the horizontal velocity for a projectile is constant, and there is no horizontal acceleration. I assume there is no horizontal acceleration, because the horizontal veloc. remains constant?
However, if there is no vertical velocity at max height then why is the object still accelerating at 9.8 m/s/s downward due to gravity? It seems like a contradiction, something can't be accelerating due to some force and not also have a velocity. Increasing velocity is the definition of acceleration, so there would be vertical velocity at max height. There is always vertical velocity due to the force of gravity accelerating the object downward.
 
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  • #2
I think the velocity at max height would just be the component 7.5 m/s in the horizontal direction

Correct. The magnitude of the the two velocity vectors are as follows

sqr[ (7.5)^2 + (0)^2 ] = 7.5

Everything in your solution looks fine to me.

However, if there is no vertical velocity at max height then why is the object still accelerating at 9.8 m/s/s downward due to gravity?

Because firstly the object the object is moving in the positive direction but accelerates towards the negative direction. Eventually its going to have to cross the point of 0m/s in order to start moving in the negative direction instead of the positive direction.

Think of a thermometer, If its 20 degrees outside and it drops to -10 degrees it eventually crossed 0 degrees right?
 
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  • #3
i understand. thanks!
 
  • #4
quick question, just to make sure i get this: the velocity for the ball at max height is the initial velocity component equation because the velocity for horizontal motion remains constant, am i right?? And the initial vertical velocity component does not apply to the first question (only with regard to finding time at max height using this equation Vy = Voy + ayt)
When i had first done this problem, i had confused the two initial component equations, not realizing this. i had tried to incorporate the vertical veloc. component equation with the first question even though i know that there is no vertical velocity. anyhow, now i know that i should have written the vertical component equation down near the second question(in my homework paper) to not confuse myself with the horizontal component equation. Both of them are found using a very similar equation. Do you see what i mean how I would confuse the two??
 
  • #5
I think i had confused the two equations because i didn't understand why the time equation
t = Vfy - Voy / -9.8 m/s/s ) uses the vertical component instead of the horizontal component. is there a reason for this??
 
  • #6
In other words, if there is no vertical velocity, why must i use a vertical veloc. component to find the time at max height. If vertical velocity is non-existent, why not use the horizontal veloc. component to find the time instead??
 
  • #7
You're thinking that there is NO vertical component of velocity and therefore its impossible to solve for time when there is NOTHING to work with right?

You need to understand when we say there is no vertical component of velocity we mean Vy = 0.

And with that fact we are able to solve for time.
 

Related to Projectile Motion at an Angle: Solving for Velocity, Time, and Height

1. What is projectile motion at an angle?

Projectile motion at an angle is the curved path that an object follows when it is thrown or launched at an angle to the ground. The object will move both horizontally and vertically, creating a parabolic trajectory.

2. What are the key factors that affect projectile motion at an angle?

The key factors that affect projectile motion at an angle are the initial velocity, the launch angle, the acceleration due to gravity, and air resistance. These factors determine the path and distance of the object's motion.

3. How does the initial velocity affect projectile motion at an angle?

The initial velocity is the speed and direction at which the object is launched. It determines the object's horizontal and vertical velocity components, which ultimately affect the shape and length of the trajectory. The greater the initial velocity, the farther the object will travel.

4. What is the optimal launch angle for maximum range in projectile motion at an angle?

The optimal launch angle for maximum range in projectile motion at an angle is 45 degrees. This angle allows for the maximum horizontal displacement and also minimizes the effect of air resistance.

5. How does air resistance affect projectile motion at an angle?

Air resistance, also known as drag, is a force that opposes the motion of an object through air. It can decrease the object's horizontal and vertical velocity, causing it to fall short of its expected range. The effect of air resistance is greater at higher launch angles and velocities.

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